Termination w.r.t. Q proof of TRS/secret06sumList.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
P(s(s(x))) → P(s(x))
SUMLIST(xs, y) → P(head(xs))
SUMLIST(xs, y) → HEAD(xs)
SUMLIST(xs, y) → TAIL(xs)
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
SUMLIST(xs, y) → INC(y)
SUMLIST(xs, y) → ISZERO(head(xs))
SUMLIST(xs, y) → ISEMPTY(xs)
SUM(xs) → SUMLIST(xs, 0)
INC(s(x)) → INC(x)
SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
P(s(s(x))) → P(s(x))
SUMLIST(xs, y) → P(head(xs))
SUMLIST(xs, y) → HEAD(xs)
SUMLIST(xs, y) → TAIL(xs)
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
SUMLIST(xs, y) → INC(y)
SUMLIST(xs, y) → ISZERO(head(xs))
SUMLIST(xs, y) → ISEMPTY(xs)
SUM(xs) → SUMLIST(xs, 0)
INC(s(x)) → INC(x)
SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

SUMLIST(xs, y) → P(head(xs))
SUMLIST(xs, y) → TAIL(xs)
SUMLIST(xs, y) → INC(y)
SUMLIST(xs, y) → ISEMPTY(xs)
SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
P(s(s(x))) → P(s(x))
SUMLIST(xs, y) → HEAD(xs)
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
SUMLIST(xs, y) → ISZERO(head(xs))
SUM(xs) → SUMLIST(xs, 0)
INC(s(x)) → INC(x)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 7 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

INC(s(x)) → INC(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
INC(x1) = INC(x1)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

[INC₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

P(s(s(x))) → P(s(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
P(x1) = P(x1)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

[P₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IF(false, false, y, xs, ys, x) → SUMLIST(ys, x)
IF(false, true, y, xs, ys, x) → SUMLIST(xs, y)
SUMLIST(xs, y) → IF(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))

The TRS R consists of the following rules:

isEmpty(cons(x, xs)) → false
isEmpty(nil) → true
isZero(0) → true
isZero(s(x)) → false
head(cons(x, xs)) → x
tail(cons(x, xs)) → xs
tail(nil) → nil
p(s(s(x))) → s(p(s(x)))
p(s(0)) → 0
p(0) → 0
inc(s(x)) → s(inc(x))
inc(0) → s(0)
sumList(xs, y) → if(isEmpty(xs), isZero(head(xs)), y, tail(xs), cons(p(head(xs)), tail(xs)), inc(y))
if(true, b, y, xs, ys, x) → y
if(false, true, y, xs, ys, x) → sumList(xs, y)
if(false, false, y, xs, ys, x) → sumList(ys, x)
sum(xs) → sumList(xs, 0)

The set Q consists of the following terms:

isEmpty(cons(x0, x1))
isEmpty(nil)
isZero(0)
isZero(s(x0))
head(cons(x0, x1))
tail(cons(x0, x1))
tail(nil)
p(s(s(x0)))
p(s(0))
p(0)
inc(s(x0))
inc(0)
sumList(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, true, x0, x1, x2, x3)
if(false, false, x0, x1, x2, x3)
sum(x0)

We have to consider all minimal (P,Q,R)-chains.